Testing Frequencies

Introduction to Research Design and Statistics

Testing Frequencies

Chi-Square Distribution

Chi-Square Statistic

df = Rows - 1

Use Table of Chi-Square

Hypotheses

H₀: The distribution of observed frequencies equals the distribution of expected frequencies.

H₁: The distribution of observed frequencies does not equal the distribution of expected frequencies.

Assumptions

Observations are independent (each subject can appear once and only once in a table)

Expected frequencies in each row are at least 15.

Example 1: Pepsi Challenge

Test whether cola preference among 220 college students in a simple random sample is equally distributed.

Each individual tastes each of the three colas. Between tastes subjects eat a soda cracker. Each subject receives the colas in a different order. Each subject then selects which soda he/she likes best.

Results: Pepsi 85, Coke 57, RC 78.

O E O-E (O-E)² (O-E)²/E

Pepsi 85 73.33 11.67 136.19 1.86
Coke 57 73.33 -16.33 266.67 3.64
RC 78 73.33 4.67 21.81 0.3
Totals 220 219.99 χ² = 5.8

	O	E	O-E	(O-E)²	(O-E)²/E
Pepsi	85	73.33	11.67	136.19	1.86
Coke	57	73.33	-16.33	266.67	3.64
RC	78	73.33	4.67	21.81	0.3
Totals	220	219.99		χ² =	5.8

df = rows - 1 = 3 - 1 = 2.

Critical value of χ² = 5.99 at alpha = 0.05.

Observed value of χ² = 5.8.

Decision: Fail to reject H₀.

Using Stata

The trick here is to create two variables, o, for the observed frequencies and e, for the expected frequencies. Then using the command chitest written by Nick Cox, which can be downloaded over the Internet (findit chitest then click to install tab_chi) to do the chi-square computation.

clear

input o
85
57
78
end

generate e = 220/3

chitest o e

observed frequencies from o; expected frequencies from e

         Pearson chi2(2) =   5.7909   Pr =  0.055
likelihood-ratio chi2(2) =   5.9984   Pr =  0.050

  +-------------------------------------------+
  | observed   expected   obs - exp   Pearson |
  |-------------------------------------------|
  |       85     73.333      11.667     1.362 |
  |       57     73.333     -16.333    -1.907 |
  |       78     73.333       4.667     0.545 |
  +-------------------------------------------+

Example 2: School Absences

A school district in a large surburban area is trying to determine if student absences are distibuted equally throughout the school week. Choosing a week at random, a researcher surveys all the schools in the district and comes up with the following data: Monday -- 73; Tuesday -- 57; Wednesday -- 48; Thusday -- 59; and Fiday -- 68.

She wants to test the absences at the 0.1 significance level.

O E O-E (O-E)² (O-E)²/E

Mon 73 61 12 144 2.4
Tue 57 61 -4 16 0.3
Wed 48 61 -13 169 2.8
Thu 59 61 -2 4 0.1
Fri 68 61 7 49 0.8
Totals 305 305 χ² = 6.4

	O	E	O-E	(O-E)²	(O-E)²/E
Mon	73	61	12	144	2.4
Tue	57	61	-4	16	0.3
Wed	48	61	-13	169	2.8
Thu	59	61	-2	4	0.1
Fri	68	61	7	49	0.8
Totals	305	305		χ² =	6.4

df = rows - 1 = 5 - 1 = 4.

Critical value of χ² = 7.78 at alpha = 0.10.

Observed value of χ² = 6.4.

Decision: Fail to reject H₀.

Using Stata

Same as pevious example

clear

input o
73
57
48
59
68
end

sum o

    Variable |     Obs        Mean   Std. Dev.       Min        Max
-------------+-----------------------------------------------------
           o |       5          61    9.77241         48         73


generate e = r(sum)/5

chitest o e

observed frequencies from o; expected frequencies from e

         Pearson chi2(4) =   6.2623   Pr =  0.180
likelihood-ratio chi2(4) =   6.3196   Pr =  0.177

  +-------------------------------------------+
  | observed   expected   obs - exp   Pearson |
  |-------------------------------------------|
  |       73     61.000      12.000     1.536 |
  |       57     61.000      -4.000    -0.512 |
  |       48     61.000     -13.000    -1.664 |
  |       59     61.000      -2.000    -0.256 |
  |       68     61.000       7.000     0.896 |
  +-------------------------------------------+

 chitable 4

        Critical Values of Chi-square
 df     .50     .25     .10     .05    .025      .01     .001
  4    3.36    5.39    7.78    9.49   11.14    13.28    18.47

Example 3: Demographics

In 1980, a Northern California county was found to be 52% anglo, 28% Hispanic, 12% African-American and 8% Asian

An SRS of 1220 individuals in 1995 found 585 Anglo, 390 Hispanic, 109 African-American and 136 Asian.

Have the demographics in the county changed greater then would be expected by chance

O E O-E (O-E)² (O-E)²/E

An 585 634.4 49.4 2,440.36 3.85
H 390 341.6 48.4 2,342.56 6.86
AA 109 146.4 -37.4 1,398.76 9.55
As 136 97.6 38.4 1,474.56 15.12
Totals 1220 1220 χ² = 35.38

	O	E	O-E	(O-E)²	(O-E)²/E
An	585	634.4	49.4	2,440.36	3.85
H	390	341.6	48.4	2,342.56	6.86
AA	109	146.4	-37.4	1,398.76	9.55
As	136	97.6	38.4	1,474.56	15.12
Totals	1220	1220		χ² =	35.38

df = rows - 1 = 4 - 1 = 3.

Critical value of χ² = 7.81 at alpha = 0.05.

Observed value of χ² = 35.38.

Decision: Reject H₀.

Using Stata

clear

input o
585
390
109
136
end

input e
634.4
341.6
146.4
97.6
end

list

     +-------------+
     |   o       e |
     |-------------|
  1. | 585   634.4 |
  2. | 390   341.6 |
  3. | 109   146.4 |
  4. | 136    97.6 |
     +-------------+

chitest o e

observed frequencies from o; expected frequencies from e

         Pearson chi2(3) =  35.3669   Pr =  0.000
likelihood-ratio chi2(3) =  34.4401   Pr =  0.000

  +-------------------------------------------+
  | observed   expected   obs - exp   Pearson |
  |-------------------------------------------|
  |      585    634.400     -49.400    -1.961 |
  |      390    341.600      48.400     2.619 |
  |      109    146.400     -37.400    -3.091 |
  |      136     97.600      38.400     3.887 |
  +-------------------------------------------+

More Stata

chitest will generate equal expected frequencies if you don't include the expected variable.

Now let's use the hsb2 dataset and the variable race.

use http://www.philender.com/courses/data/hsb2, clear

chitest race, count  /* down loaded from the Internet */

Chi-square test:
    observed frequencies from race
    expected frequencies equal

         Pearson chi2(3) = 242.4400   Pr =  0.000
likelihood-ratio chi2(3) = 203.5732   Pr =  0.000

                                   residuals
      observed    expected     classic     Pearson 
  1.        24      50.000     -26.000      -3.677  
  2.        11      50.000     -39.000      -5.515  
  3.        20      50.000     -30.000      -4.243  
  4.       145      50.000      95.000      13.435

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Phil Ender, 28nov05, 22Nov00